1,5

Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g. if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11, 2001.

Julius Worpitzky's 1883 algorithm generates Bernoulli numbers. By way of example [Wikipedia]: B0 = 1 B1 = 1/1 - 1/2 B2 = 1/1 - 3/2 + 2/3 B3 = 1/1 - 7/2 + 12/3 - 6/4 B4 = 1/1 - 15/2 + 50/3 - 60/4 + 24/5 B5 = 1/1 - 31/2 + 180/3 - 390/4 + 360/5 - 120/6 B6 = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7 ...(note that in this algorithm odd n's for the Bernoulli numbers sum to 0, not 1 and the sum for B1 = 1/2 = (1/1 - 1/2). B3 = 0 =(1 - 7/2 + 13/3 - 6/4) = 0. The summation for B4 = -1/30. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 09 2008]

Contribution from Tom Copeland (tcjpn(AT)msn.com), Oct 23 2008: (Start)

G(x,t) = 1/ {1 + [1-exp(x t)]/t} = 1 + 1 x + (2 + t) x^2/2! + (6 + 6t + t^2) x^3/3! + ...

gives row polynomials for A090582-- reverse f-polynomials for the permutohedra (see A019538).

G(x,t-1) = 1 + 1 x + (1 + t) x^2 / 2! + (1 + 4t + t^2) x^3 / 3! + ...

gives row polynomials for A008292, the h-polynomials for permutohedra.

G[(t+1)x,-1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ...

gives row polynomials for A028246.

(End)

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 18 2009: (Start)

The Worpitzky triangle seems to be an apt name for this triangle.

(End)

A. Riskin and D. Beckwith, Problem 10231, Amer. Math. Monthly, 102 (1995), 175-176.

Wikipedia (Bernoulli numbers). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 09 2008]

H. Hasse, Ein Summierungsverfahren fuer die Riemannsche Zeta-Reihe.

N. J. A. Sloane, Transforms

E.g.f.: -ln(1-y*(exp(x)-1)). - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 28 2003

a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.

Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deleham's operator defined in A084938, but the notation is different.

Sum of terms in n-th row = A000629(n) - Gary W. Adamson (qntmpkt(AT)yahoo.com), May 30 2005

The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t)=t(t+1)diff(P(n, t), t) for n>=1 (see the Riskin and Beckwith reference). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 09 2005

Additional comments from Gottfried Helms, Jul 12 2006:

(Start) Delta-matrix as can be read from H. Hasse's proof of a connection

between the zeta-function and Bernoulli numbers (see link below).

Let P = lower triangular matrix with entries P[row,col] = binom(row,col)

Let J = unit matrix with alternating signs J[r,r]=(-1)^r

Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers

Let V = Vandermonde matrix with V[r,c] = (r+1)^c

V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0)

Then Delta = P*J * V and B' = N(-1)' * Delta

where B is the column matrix of Bernoulli numbers and ' means transpose,

or for the single k'th Bernoulli number B_k with the appropriate column of Delta

B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k)

Using a single column instead of V and assuming infinite dimension

H. Hasse showed that in x = N(-1) * P*J * N(s),

where s can be any complex number and s*zeta(1-s) = x.

His theorem reads: s*zeta(1-s) = sum_{n=0..inf} ( (n+1)^-1 * delta(n,s) )

where delta(n,s) = sum_{j=0..n} [ (-1)^j * binom(n,j) * (j+1)^s ] (end)

The k-th row (k>=1) contains a(i, k) for i=1 to k, where a(i, k) satisfies Sum_{i=1..n} C(i, 1)^k = 2 * C(n+1, 2) * Sum_{i=1..k} a(i, k) * C(n-1, i-1)/(i+1). E.g. Row 3 contains 1, 3, 2 so Sum_{i=1..n} C(i, 1)^3 = 2 * C(n+1, 2) * [ a(1, 3)/2 +a(2, 3) *C(n-1, 1)/3 +a(3, 3)*C(n-1, 2)/4 ] = [ (n+1)*n ] * [ 1/2 +(3/3)*C(n-1, 1) +(2/4)*C(n-1, 2) ] = ( n^2 +n ) * ( n -1 +[ C(n-1, 2) +1 ]/2 ) = C(n+1, 2)^2. See A000537 for more details ( 1^3 +2^3 +3^3 +4^3 +5^3 +... ). - Andre F. Labossiere (boronali(AT)laposte.net), Sep 22 2003

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 18 2009: (Start)

a(n,k) = k*a(n-1,k) + (k-1)*a(n-1,k-1) with a(n,1) = 1 and a(n,n) = (n-1)!

(End)

1; 1,1; 1,3,2; 1,7,12,6; 1,15,50,60,24; ...

Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.

a:=(n, k)->add( (-1)^(k-i)*C(k, i)*i^n, i=0..k)/k;

(PARI) T(n, k)=if(k<0|k>n, 0, n!*polcoeff((x/log(1+x+x^2*O(x^n)))^(n+1), n-k))

Dropping the column of 1's gives A053440. See also A008277.

Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.

Cf. A087127, A087107, A087108, A087109, A087110, A087111, A084938 A075263.

Row sums give A000629.

A027642, A002445 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 09 2008]

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 18 2009: (Start)

Appears in A161739 (RSEG2 triangle), A161742 and A161743.

(End)

Sequence in context: A056151 A134436 A163626 this_sequence A082038 A143774 A158474

Adjacent sequences: A028243 A028244 A028245 this_sequence A028247 A028248 A028249

nonn,easy,nice,tabl

N. J. A. Sloane (njas(AT)research.att.com), Doug McKenzie (mckfam4(AT)aol.com)

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Jan 14 2000

Definition corrected by Li Guo, Dec 16 2006

Link to the Hasse paper repaired Peter Luschny (peter(AT)luschny.de), Apr 21 2009

Typo in link corrected by Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 17 2009